Question

A wedge of mass $m_3$ is placed on a smooth horizontal surface. Blocks of mass $m_1\text{and}{m}_2$ are placed on the smooth inclined faces of $m_3$ as shown. If $m_3$ does not move, the ratio $\frac{m_2}{m_1}$ is equal to

Answer 1

$m_3$ will be at rest if component of $N_1$ along horizontal is balanced by component of $N_2$ along horizontal.
$N_1\cos {45}^{\circ }=N_2\cos {75}^{\circ }$
$\left(m_{1}g\cos {45}^{\circ }\right)\cos {45}^{\circ }=m_{2}g\cos {15}^{\circ }\cos {75}^{\circ }$
$\Rightarrow \frac{m_2}{m_1}=\frac{{\cos }^2{45}^{\circ }}{\cos {15}^{\circ }\cos {75}^{\circ }}$
$\Rightarrow \frac{m_2}{m_1}=\frac{\frac{1}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}\times \frac{\sqrt{3}-1}{2\sqrt{2}}}$
$\Rightarrow \frac{m_2}{m_1}=\frac{8}{4}=2$